Formation of the Cu -Phenoxyl Radical by Reaction of O with a Cu -Phenolate Complex via the Cu -Phenoxyl Radical.

Reaction of Cu(ClO ) ⋅6 H O with a tripodal 2N2O ligand, H Me NL, having a p-(dimethylamino)phenol moiety, in CH Cl /MeOH (1:1 v/v) under basic conditions under an inert gas atmosphere gave [Cu(Me NL)(H O)] (1). The same reaction carried out under aerobic conditions gave [Cu(Me NL)(MeOH)]ClO (2), which could be obtained also from the isolated complex 1 by reaction with O in CH Cl /MeOH. The X-ray crystal structures of 1 and 2 revealed similar square-pyramidal structures, but 2 showed the (dimethylamino)phenoxyl radical features. Complex 1 exhibits characteristic Cu EPR signals of the d ground state in CH Cl /MeOH at 77 K, whereas 2 is EPR-silent. The EPR and X-ray absorption fine structure (XAFS) results suggest that 2 is assigned to the Cu -(dimethylamino)phenoxyl radical. However, complex 1 showed different features in the absence of MeOH. The EPR spectrum of the CH Cl solution of 1 exhibits distortion from the d ground state and a temperature-dependent equilibrium between the Cu -(dimethylamino)phenolate and the Cu -(dimethylamino)phenoxyl radical. From these results, Cu -phenoxyl radical complex 2 is concluded to be formed by the reaction of 1 with O via the Cu -phenoxyl radical species.